std::ranges::rotate_copy, std::ranges::rotate_copy_result - cppreference.com (original) (raw)

Defined in header
Call signature
template< std::forward_iterator I, std::sentinel_for<I> S, std::weakly_incrementable O >requires std::indirectly_copyable<I, O> constexpr rotate_copy_result<I, O> rotate_copy( I first, I middle, S last, O result ); (1) (since C++20)
template< ranges::forward_range R, std::weakly_incrementable O > requires std::indirectly_copyable<ranges::iterator_t<R>, O> constexpr rotate_copy_result<ranges::borrowed_iterator_t<R>, O> rotate_copy( R&& r, ranges::iterator_t<R> middle, O result ); (2) (since C++20)
Helper types
template< class I, class O > using rotate_copy_result = in_out_result<I, O>; (3) (since C++20)

Copies the left rotation of [first, last) to result.

  1. Copies the elements from the source range [first, last), such that in the destination range, the elements in [first, middle) are placed after the elements in [middle, last) while the orders of the elements in both ranges are preserved.

The behavior is undefined if either [first, middle) or [middle, last) is not a valid range, or the source and destination ranges overlap.

  1. Same as (1), but uses r as the source range, as if using ranges::begin(r) as first and ranges::end(r) as last.

The function-like entities described on this page are algorithm function objects (informally known as niebloids), that is:

Contents

[edit] Parameters

first, last - the iterator-sentinel pair defining the source range of elements to copy from
r - the source range of elements to copy from
middle - the iterator to the element that should appear at the beginning of the destination range
result - beginning of the destination range

[edit] Return value

{last, result + N}, where N = ranges::distance(first, last).

[edit] Complexity

Linear: exactly N assignments.

[edit] Notes

If the value type is TriviallyCopyable and the iterator types satisfy contiguous_iterator, implementations of ranges::rotate_copy usually avoid multiple assignments by using a "bulk copy" function such as std::memmove.

[edit] Possible implementation

See also the implementations in libstdc++ and MSVC STL.

struct rotate_copy_fn { template<std::forward_iterator I, std::sentinel_for S, std::weakly_incrementable O> requires std::indirectly_copyable<I, O> constexpr ranges::rotate_copy_result<I, O> operator()(I first, I middle, S last, O result) const { auto c1 {ranges::copy(middle, std::move(last), std::move(result))}; auto c2 {ranges::copy(std::move(first), std::move(middle), std::move(c1.out))}; return {std::move(c1.in), std::move(c2.out)}; }   template<ranges::forward_range R, std::weakly_incrementable O> requires std::indirectly_copyable<ranges::iterator_t, O> constexpr ranges::rotate_copy_result<ranges::borrowed_iterator_t, O> operator()(R&& r, ranges::iterator_t middle, O result) const { return (*this)(ranges::begin(r), std::move(middle), ranges::end(r), std::move(result)); } };   inline constexpr rotate_copy_fn rotate_copy {};

[edit] Example

#include #include #include #include   int main() { std::vector src {1, 2, 3, 4, 5}; std::vector dest(src.size()); auto pivot = std::ranges::find(src, 3);   std::ranges::rotate_copy(src, pivot, dest.begin()); for (int i : dest) std::cout << i << ' '; std::cout << '\n';   // copy the rotation result directly to the std::cout pivot = std::ranges::find(dest, 1); std::ranges::rotate_copy(dest, pivot, std::ostream_iterator(std::cout, " ")); std::cout << '\n'; }

Output:

[edit] See also